Question

这应该列出用户定义的文本文件中的每个单词，然后将它们放入一个数组中。这是我长期以来第一次尝试java，我只知道这是一团糟。

public static String[] parse(String path) throws IOException {
    List<String> list = new ArrayList<String>();
    String line;
    try <BufferedReader br = new BufferedReader(new FileReader(path))) {
        while((line = br.readLine()) != null) {
        // Trying to make sure I make words between the spaces and punctuation

            for (int i = 0; i < line.length(); i++) {
                int x;
                if (Character.isLetter(line.charAt(i))) {
                    x = i;}
                    for (int a = x; a < line.length(); a++) {
                        if (!Character.isLetter(line.charAt(a))) {
                            String s = line.substring(x, a)
                            list.add(s);
                            i = a;
                            a = line.length() + 1;
                        }
                    }
                }
            }
        }
        String[] arr = list.toArray(new String[list.size()]);
        return arr;
    }
    catch (Exception e) {
        System.out.println("could not find file");
        return null;
    }
}

Answer 1

如果要求从文件中获取所有单词并将其放入列表中，您可以通过以下更简单的方式完成此操作：

List<MyModel> data = (List<MyModel>)TempData.Peek("data");

你的清单现在可以包含所有单词。将列表转换为字符串[] ..

while ((line = br.readLine()) != null) {
    String[] words = line.split(" ");
     // Now you have a String array containing each word in the current line
     //Add all words to list as below.
     list.addAll(Arrays.asList(words));
   }

Answer 2

我更喜欢拆分一个或多个空格字符（修剪线后），然后使用正则表达式删除所有字母而不是字母（而不是字符测试）。你可以在List.toArray中使用一个空数组，它会为你调整大小。永远不要默默地吞下异常。像，

public static String[] parse(String path) throws IOException {
    List<String> list = new ArrayList<>();
    String line;
    try (BufferedReader br = new BufferedReader(new FileReader(path))) {
        while ((line = br.readLine()) != null) {
            String[] tokens = line.trim().split("\\s+");
            for (String token : tokens) {
                token = token.replaceAll("[^a-zA-Z]", "");
                if (!token.isEmpty()) {
                    list.add(token);
                }
            }
        }
        return list.toArray(new String[0]);
    } catch (Exception e) {
        System.out.println("could not find file " + e.getMessage());
        e.printStackTrace();
        return null;
    }
}

Answer 3

在这种情况下，使用Scanner更合适，速度更快：

List<String> words = new ArrayList<>();
try (Scanner sc = new Scanner(new File(path))) {
    //Everything that's not a character is treated as delimiter
    sc.useDelimiter("[^a-zA-Z]+");
    while (sc.hasNext()) {
        words.add(sc.next());
    }
}

对于Java 9来说，它甚至更简单：

final List<String> words;
try (Scanner sc = new Scanner(new File(path))) {
    sc.useDelimiter("[^a-zA-Z]+");
    words = sc.tokens().collect(Collectors.toList());
}

有没有更好的方法来编写这个Java方法？

3 个答案: